For `PCl_(5)hArr PCl_(3)+Cl_(2)`, initial concentration of each reactant and product is 1 M. If `K_(eq)=0.41` then

Options

A. More `PCl_(3)` will form
B. More `Cl_(2)` will form
C. More `PCl_(5)` will form
D. No change

Answer

To solve the problem regarding the equilibrium of the reaction \( PCl_5 \rightleftharpoons PCl_3 + Cl_2 \) with given initial concentrations and equilibrium constant, we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_{eq} \).
The equilibrium constant expression for the reaction is given by:

\[ K_{eq} = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 2: Identify the initial concentrations.
According to the problem, the initial concentrations of \( PCl_5 \), \( PCl_3 \), and \( Cl_2 \) are all 1 M:

\[ [PCl_5]_0 = 1 \, M, \quad [PCl_3]_0 = 1 \, M, \quad [Cl_2]_0 = 1 \, M \]

### Step 3: Determine the change in concentrations at equilibrium.
Let \( x \) be the amount of \( PCl_5 \) that dissociates at equilibrium. Therefore, at equilibrium, the concentrations will be:

\[ [PCl_5] = 1 - x \]
\[ [PCl_3] = 1 + x \]
\[ [Cl_2] = 1 + x \]

### Step 4: Substitute the equilibrium concentrations into the \( K_{eq} \) expression.
Substituting these expressions into the equilibrium constant equation gives:

\[ K_{eq} = \frac{(1 + x)(1 + x)}{(1 - x)} \]

### Step 5: Set up the equation with the given \( K_{eq} \).
We know that \( K_{eq} = 0.41 \). Therefore, we can set up the equation:

\[ 0.41 = \frac{(1 + x)^2}{(1 - x)} \]

### Step 6: Cross-multiply and simplify the equation.
Cross-multiplying gives:

\[ 0.41(1 - x) = (1 + x)^2 \]

Expanding both sides:

\[ 0.41 - 0.41x = 1 + 2x + x^2 \]

Rearranging the equation:

\[ x^2 + (2 + 0.41)x + (1 - 0.41) = 0 \]
\[ x^2 + 2.41x + 0.59 = 0 \]

### Step 7: Solve the quadratic equation for \( x \).
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ x = \frac{-2.41 \pm \sqrt{(2.41)^2 - 4 \cdot 1 \cdot 0.59}}{2 \cdot 1} \]
\[ x = \frac{-2.41 \pm \sqrt{5.8081 - 2.36}}{2} \]
\[ x = \frac{-2.41 \pm \sqrt{3.4481}}{2} \]
\[ x = \frac{-2.41 \pm 1.86}{2} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{-0.55}{2} = -0.275 \) (not possible since concentration cannot be negative)
2. \( x = \frac{-4.27}{2} = -2.135 \) (not possible)

### Step 8: Analyze the result.
Since both potential values for \( x \) are negative, this indicates that the reaction will shift to the left, favoring the formation of \( PCl_5 \).

### Conclusion:
Since \( K_{eq} < Q_c \), the equilibrium will shift towards the reactants, resulting in an increase in the concentration of \( PCl_5 \) and a decrease in the concentrations of \( PCl_3 \) and \( Cl_2 \).

### Final Answer:
The correct conclusion is that more \( PCl_5 \) will be formed at equilibrium.

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